SpinSpin--spin coupling analysisspin coupling analysis

• The last parameter that we will discuss concerning the interpretation of NMR spectra is the 1H spin-spin coupling. Couplings are perhaps the most important parameter in NMR, as they allow us to elucidate chemical structure.

• Scalar spin-spin coupling shows up as a splitting, or fine structure, in our spectrum. It will occur between two magnetically active nuclei that are connected through chemical bonds. We can see it for two atoms directly bonded,or for atoms that ‘see’ one another across several bonds.

J ≠ 0J = 0

• A splitting of a signal means that we have more energies involved in the transition of a certain nuclei. So why do we have more energies?

• The reason is the bonding electrons and their magnetic moments. The magnetic moment of the nuclei produces a small polarization (orientation…) of the bonding electron, and this is transmitted by overlapping orbitals to the other nuclei.

I SI S

SpinSpin--spin coupling (continued)spin coupling (continued)

• We can explain this better by looking at HF:

Bo

19F

19F

1H

1H

Nucleus

Electron

• The nuclear magnetic moment of 19F polarizes the F bonding electron (up), which, since we are following quantum mechanics rules, makes the other electron point down (the electron spins have to be antiparallel).

• Now, the since we have different states for the 1�H electrons depending on the state of the 19F nucleus, we will have slightly different energies for the 1H nuclear magnetic moment (remember that the 1s electron of the 1H generates an induced field…).

• This difference in energies for the 1H result in a splitting of the 1H resonance line.

SpinSpin--spin coupling (…)spin coupling (…)

• We can do a similar analysis for a CH2 group:

Bo

1H 1H

C

• The only difference here is that the C bonds are hybrid bonds (sp 3), and therefore the Pauli principle and Hundi’s rulespredict that the two electrons will be parallel.

• Irrespective of this, the state of one of the 1H nuclei is transmitted along the bonds to the other 1H, and we get a splitting (a doublet in this case…). The energy of the interactions between two spins A and B can be found by therelationship:

E = JAB * IA * IB

SpinSpin--spin coupling (…)spin coupling (…)

• IA and IB are the nuclear spin vectors, and are proportional toµµµµA and µµµµB, the magnetic moments of the two nuclei. JAB is the scalar coupling constant . So we see a very important feature of couplings. It does not matter if we have a 60, a 400, or an 800 MHz magnet, the coupling constants arealways the same!!!

• Lets do a more detailed analysis in term of the energies. Lets think a two energy level system, and the transitions for nuclei A. When we have no coupling (J = 0), the energy involved in either transition (A1 or A2) is equal (no spin-spin interaction).

A XA X

J = 0J > 0

• The relative orientations of the nuclear moments does notmatter. We see a single line (two with equal frequency).

• When J > 0, the energy levels of the spin system will be either stabilized or destabilized. Depending on the relative orientations of the nuclear moments, the energies for the A1and A2 transition will change giving two different frequencies (two peaks for A).

A X

A1

A2

A1

A2

Bo E

J = 0E4

E3E2

E1

SpinSpin--spin coupling (…)spin coupling (…)

• We choose J > 0 as that related to antiparallel nuclear moments (lower energy). The energy diagram for J < 0 would then be:

A X A X

A1

A2

A1

A2

Bo E

J = 0 J < 0E4

E3E2

E1

• If we look at it as a stick spectrum for either case we get:

• As mentioned before, the choice of positive or negative J is a definition. However, we see that we won’t be able to tell if we have a positive or negative J, because the lines in the spectrum corresponding to the different transitions basically change places. Unless we are interested in studying the energies, this is not important for structure elucidation…

A1

A2

νννν A1 = νννν A2

A2 A1

νννν A2 νννν A1

A1 A2

νννν A1 νννν A2

J = 0J > 0 J < 0

SpinSpin--spin coupling (…)spin coupling (…)

• We can do a quantitative analysis of the energy values from to gain some more insight on the phenomenon. The base energies of the system are related to the Larmorfrequencies, and the spin-spin interaction is JAX:

A1

A2

J > 0

I2

I1

E4 = 1/2 ννννA + 1/2 ννννB + 1/4 JAX

E3 = 1/2 ννννA - 1/2 ννννB - 1/4 JAX

E2 = - 1/2 ννννA + 1/2 ννννB - 1/4 JAX

E1 = - 1/2 ννννA - 1/2 ννννB + 1/4 JAX1

23

4

• So if we now consider the transitions that we see in the spectrum, we get:

• This explains the lines in our spectrum quantitatively…

E1 = - /2 ννννA - /2 ννννB + /4 JAX1

A1 = E4 - E2 = ννννA - 1/2 JAX

A2 = E3 - E1 = ννννA + 1/2 JAX

I1 = E2 - E1 = ννννB - 1/2 JAX

I2 = E4 - E3 = ννννB + 1/2 JAX

Analysis of 1Analysis of 1stst order systemsorder systems

• Now we will focus on the simplest type of coupling we can have, which is one of the limits of a more complex quantum mechanical description.

• Lets say that we have ethylacetate. In this molecule, the resonance of the CH3 from the ethyl will be ~ 1.5, while that for the CH2 will be ~ 4.5 ppm. In most spectrometers this means that the difference in chemicals shifts, ∆ν∆ν∆ν∆ν, will be a lot bigger than the coupling constant J, which for this system is ~ 7 Hz. In this case, we say that we have a first order spin system .

• If we analyze the system in the same way we did the simple AX system, we will see that each 1H on the CH2 will see 4 possible states of the CH3

1Hs, while each 1H on the CH3will see 3 possible states of the CH2 protons. We have to keep in mind that the two 1Hs of the CH2 and the three 1Hs of the CH3 are equivalent.

• In order to see this better, we can build a diagram that has the possible states of each 1H type in EtOAC:

αααααααααβ βααβ βααβ βααβ βα

ββββββββ

ααααααααααααααβ αβα βααααβ αβα βααααβ αβα βααααβ αβα βαααββ βαβ ββααββ βαβ ββααββ βαβ ββααββ βαβ ββα

ββββββββββββCH3 CH2

11stst order systems (continued)order systems (continued)

• If we generalize, we see that if a certain nuclei A is coupled to n identical nuclei X (of spin 1/2), A will show up as n + 1lines in the spectrum. Therefore, the CH2 in EtOAc will show up as four lines, or a quartet . Analogously, the CH3 in EtOAc will show up as three lines, or a triplet .

• The separation of the lines will be equal to the coupling constant between the two types of nuclei (CH2’s and CH3’s in EtOAc, approximately 7 Hz).

• If we consider the diagram of the possible states of each nuclei, we can also see what will be the intensities of the lines:

• Since we have the same probability of finding the system in any of the states, and states in the same rows have equal energy, the intensity will havea ratio 1:2:1 for the CH3, anda ratio of 1:3:3:1 for the CH2:

CH3 CH2

CH3

CH2

J (Hz)

4.5 ppm 1.5 ppm

11stst order systems (…)order systems (…)

• These rules are actually a generalization of a more general rule. The splitting of the resonance of a nuclei A by a nuclei X with spin number I will be 2I + 1.

• Therefore, if we consider -CH2-CH3,and the effect of each of the CH3’s1Hs, and an initial intensity of 8,we have:

8

4 4Coupling to the first 1H(2 * 1/2 + 1 = 2)

• Since the coupling to each 1H of the CH3 is the same, the lines will fall on top of one another.

• In general, the number of lines in these cases will be a binomial expansion, known as the Pascal Triangle :

2 22 2

1 111 1111

Coupling to the second 1H

Coupling to the third 1H

1 : n / 1 : n ( n - 1 ) / 2 : n ( n - 1 ) ( n - 2 ) / 6 : ...

11stst order systems (…)order systems (…)

• Here n is the number of equivalentspins 1/2 we are coupled to: Theresults for several n’s:

• In a spin system in which we have a certain nuclei coupled to more than one nuclei, all first order, the splitting will be basically an extension of what we saw before.

• Say that we have a CH (A) coupled to a CH3 (M) with a JAMof 7 Hz, and to a CH2 (X) with a JAX of 5 Hz. We basically go in steps. First the big coupling, which will give a quartet:

11 1

1 2 11 3 3 1

1 4 6 4 1

• Then the small coupling,which will split each linein the quartet into a triplet:

• This is called a triplet of quartets (big effect is the last…).

7 Hz

5 Hz

11stst order systems (…)order systems (…)

• Lets finish our analysis of 1st order system with some pretty simple rules that we can use when we are actually looking at 1D 1H spectra.

• To do that, say that againwe have a system in with aCH (A) coupled to two CH’s(M and R) with a JAM of 8 Hzand JAR of 5 Hz, and to aCH2 (X) with a JAX or 6 Hz:

8 Hz

5 Hz

• The first rule is that if we have a clear-cut first order system, the chemical shift of nuclei A, δδδδA, is always at the center of mass of the multiplet.

• The second one is that no matter how complicated the pattern may end up being, the outermost splitting is always the smallest coupling for nuclei A. We can measure it withoutworrying about picking the wrong peaks from the pattern.

6 Hz

5 Hz δδδδA

The Karplus equationThe Karplus equation

• The most common coupling constant we’ll see is the three-bond coupling, or 3J:

1H

C

C

1H

φφφφ

• As with the 1J or 2J, the coupling arises from the interactions between nuclei and electron spins. 1J and 3J will hold the same sign, while 2J will have opposite sign.

• However, the overlap of electron and nuclear wavefunctions in the case of 3J couplings will depend on the dihedral angle <φφφφ> formed between the CH vectors in the system.

• The magnitude of the 3J couplings will have a periodic variation with the torsion anlge, something that was first observed by Martin Karplus in the 1950’s.

3JHH = A · cos2( φφφφ ) + B · cos( φφφφ ) + C

The Karplus equation (...)The Karplus equation (...)

• The relationship can be expressed as a cosine series:

• Here A, B, and C are constants that depend on the topology of the bond (i.e., on the electronegativity of the substituents).Graphically, the Karplus equation looks like this:

3 JH

H(H

z)

φφφφ (o)

• A nice ‘feature’ of the Karplus equation is that we can estimate dihedral angles from 3J coupling constants. Thus, avariety of A, B, and C parameters have been determined forpeptides, sugars, etc., etc.

C

CH1

H2

φφφφ

3JHH = 1.7 Hz

The Karplus equation (...)The Karplus equation (...)

• As an example, consider an SN2 inversion reaction we do inour lab. We go from a trans to a cis β-lactam.

N

TsO

Ph

O

PMP

3JHH = 5.4 Hz

N

N3

Ph

O

PMP

• The <φφφφHH> in the trans is 137.3o, while in the cisproduct it is 1.9o. So, we know we have inversion...

N3-

22ndnd order systems order systems -- The AB systemThe AB system

• What we have been describing so far is a spin system in which ∆ν∆ν∆ν∆ν >> J, and as we said, we are analyzing one of the limiting cases that QM predict.

• As ∆ν∆ν∆ν∆ν approaches J, there will be more transitions of similar energy and thus our spectrum will start showing more signals than our simple analysis predicted. Furthermore, the intensities and positions of the lines of the multiplets will be different from what we saw so far.

• Lets say that we have twocoupled nuclei, A and B,and we start decreasingour B . ∆ν∆ν∆ν∆ν will get smaller

∆ν∆ν∆ν∆ν >> J

our Bo. ∆ν∆ν∆ν∆ν will get smallerwith J staying the same.After a while, ∆ν∆ν∆ν∆ν ~ J. Whatwe see is the following:

• What we did here is tostart with an AX system(the chemical shifts ofA and X are very different)and finish with an ABsystem, in which ∆ν∆ν∆ν∆ν ~ J.

• Our system is now a secondorder system. We have effects that are not predicted by the simple multiplicity rules that we described earlier.

∆ν∆ν∆ν∆ν = 0

The AB system (continued)The AB system (continued)

• Thus, the analysis of an AB system is not as straightforward as the analysis of an AX system. A full analysis cannot be done without math and QM, but we can describe the results.

• A very simple way to determineif we have an AB system is bylooking at the roofing effect :coupled pairs will lean towardseach other, making a little roof:

• As with an AX system, JABis the separation between

1 2 3 4

1 2 3 4

A BννννZννννA ννννB

| JAB | = | f1 - f2 | = | f3 - f4 |lines 1 and 2 or 3 and 4:

• Now, the chemical shifts of nuclei A and B are not at the center of the doublets. They will be at the center of mass of both lines. Being ∆ν∆ν∆ν∆ν the ννννA - ννννB chemical shift difference, ννννAand ννννB will be:

| JAB | = | f1 - f2 | = | f3 - f4 |

∆ν∆ν∆ν∆ν2 = | ( f1 - f4 ) ( f2 - f3 ) |

ννννA = ννννZ - ∆ν∆ν∆ν∆ν / 2

ννννB = ννννZ + ∆ν∆ν∆ν∆ν / 2I2 I3 | f1 - f4 |

= = I1 I4 | f2 - f3 |

• Peak intensities can becomputed similarly:

Magnetic and chemical equivalenceMagnetic and chemical equivalence

• Before we get deeper into analysis of coupling patterns, lets pay some more attention to naming conventions, as well as to some concepts regarding chemical and magnetic equivalence.

• Our first definition will be that of a spin system . We have a spin system when we have a group of n nuclei (with I = 1/2) that is characterized by no more than n frequencies (chemical shifts) ννννi and n ( n - 1 ) / 2 couplings J ij. The couplings have to be within nuclei in the spin system.

• We start by defining magnetic equivalence by analyzing some examples. Say that we have an ethoxy group (-O-CH2-CH ). CH3).

• As we saw last time, we can do a very simple first order analysis of this spin system, because we assumed that all CH2 protons were ‘equal,’ and all CH3 protons were ‘equal.’ Is this true?

• We can easily see that they are chemically equivalent. Additionally, we have free rotation around the bond, which makes their chemical shifts and couplings equal.

Ha

OR

HbH1

H2H3

Ha

OR

HbH3

H1H2

Ha

OR

HbH2

H3H1

Magnetic equivalence (continued)Magnetic equivalence (continued)

• Since the 1Hs can change places, they will alternate their chemical shifts (those bonded to the same carbon), and we will see an average.

• The same happens for the J couplings. We’ll see an average of all the JHH couplings, so in effect, the coupling of any proton in CH2 to any proton in the CH3 will be the same.

• If we introduce some notation, and remembering that δ(CH2) is >> δ(CH3), this would be an A2X3 system: We have 2 magnetically equivalent 1Hs on the CH2, and 3 on the CH3.

• The 2JHH coupling (that is, the coupling between two nuclei bound to the same carbon) is zero in this case, because the bound to the same carbon) is zero in this case, because the energies for any of the three (or two) protons is the same.

• Finally, we use A to refer to the CH2 protons, and X to refer to the CH3 protons because they have very different δs. We usually start with the letter A for the most deshielded spin.

• Difluoromethane is another example of an ‘AX’ type system:

• In this case, 1Hs and 19Fs are equal not due to rotation, but to symmetry around the carbon. It’s an A2X2 system.

H

HF

F

Magnetic equivalence (…)Magnetic equivalence (…)

• For CH2F2, we can also compare the couplings to check that the 1Hs and 19Fs are equivalent: JH1F1 = JH1F2 = JH2F1 = JH2F2.All due to their symmetry...

• Now, what about thethe 1Hs and 19Fs in 1,1-difluoroethene?

• Here we also have symmetry, but no rotation. The two 1Hs and the two 19Fs are chemically equivalent, and we can

δδδδHa δδδδHb δδδδFa δδδδFb

Ha

Hb Fb

Fa

easily see that δδδδHa = δδδδHb and δδδδFa = δδδδFb.

• However, due to the geometry of this compound, JHaFa ≠≠≠≠JHaFb. Analogously, JHbFa ≠≠≠≠ JHbFb.

• Furthermore, since the couplings are different, the energy levels for Ha and Hb are different (not degenerate anymore as in CH3), and we have JHaHb ≠≠≠≠ 0.

• If we consider all the possible couplings we have, we have three different couplings for each proton. For Ha, we have JHaFa, JHaFb, and JHaHb. For Hb, we have JHbFa, JHbFb, and JHbHa. This means more than the eight possible transitions (2*2*2) in the energy diagram, and an equal number of possible lines in the spectrum!

Magnetic equivalence (…)Magnetic equivalence (…)

• These are representative spectra (only the 1H spectrum is shown) of CH2F2 and F2C=CH2:

CH2F2

H2C=CF2

• A system like this is not an A2X2, but an AA’XX’ system. We have two A nuclei with the same chemical shift but that are not magnetically equivalent. The same goes for the X nuclei.

• The following are other examples of AA’XX’ systems:

X Y

HA

HA' HX'

HX

O

HA

HX HX'

HA'

Energy diagrams for 2Energy diagrams for 2ndnd order systemsorder systems

• From what we’ve seen, most cases of magnetic non-equivalence give rise to 2nd order systems, because we will have two nuclei with the same chemical environment and the same chemical shift, but with different couplings (AA’ type…).

• We have analyzed qualitatively how a 2nd order AB looks like.In an AB system we have two spins in which ∆δ∆δ∆δ∆δ ~ JAB. The energy diagram looks a lot like a 1st order AX system, but the energies involved (frequencies)and the transition probabilities(intensities) are such that weget a messier spectrum:

ββAB

• Some examples of AB systems:αα

αβ βα

A B

S

HA HB

ClBr

Br

Cl HB

HA

CHB

CHAO O

CHACHB

Transition from 1Transition from 1stst order to 2order to 2ndnd orderorder

• The following is a neat experimental example of how we go from a 1st order system to a 2nd order system. The protons in the two compounds have the same ‘arrangement’, but as ∆δ∆δ∆δ∆δ approaches JAB, we go from, in this case, A2X to A2B:

Cl

Cl

Cl

HX Cl

ClHA HA

Cl

Cl

HA

HB

HA

Cl

• Most examples have ‘ringing’ and are obtained at relatively low fields (60 MHz) because at these fields 2nd order effects are more common…

22ndnd order systems with more than 2 spins.order systems with more than 2 spins.

• Now, lets analyze 2nd order systems with more than 2 spins. We already saw an example, the A2X system and the A2Bsystem. The A2X system is 1st order, and is therefore easy to analyze.

• The A2B is 2nd order, and energy levels includes transitions for what are known as symmetric and antisymmetric wavefunctions . They are related with the symmetry of the quantum mechanical wavefunctions describing the system.

• In any case, we have additional transitions from the ones we see in a A2X system:

A XA2B

• We now have 9 lines in the A2B system, instead of the 5 we have in the A2X system (we have to remember that in the A2X many of the transitions are of equivalent energy…).

βββ

ββα βαβ αββ

βαα αβα ααβ

ααα

βββ

β(αβ+βα) αββ β(αβ−βα)

βαα α(αβ+βα) α(αβ−βα)

ααα

A2XA2B

symmetric

antisymmetric

More than 2 spins (continued)More than 2 spins (continued)

• An A2B (or AB2) spectrumwill look like this:

OH

HO

HA

HB

HA

OH

• Another system that we will encounter is the ABX system, in which two nuclei have comparable chemical shifts and a third is far from them. The energy levels look like this:

ββ

βα αβ

αβ

ββ

βα αβ

αα

X(αααα)

X(ββββ)

A

A

A

AB

B

B

B

More than 2 spins (…)More than 2 spins (…)

• In a ABX spectrum, we will have 4 lines for the A part, 4 lines for the B part, and 6 lines for the X part:

O

HA

HBHX

Br

• ABX spin systems are very common in trisubstituted aromatic systems.

• The last system we will discuss is the AA’BB’ system, that we saw briefly at the beginning (actually, we saw AA’XX’ …).

More than 2 spins (…)More than 2 spins (…)

• In an AA’BB’/AA’XX’ system we have 2 pairs of magnetically non-equivalent protons with the same chemical shift. The energy diagram for such as system is:

ββββ

αααβ ααβα

ααββ αβαβ βααβ βαβα

αβββ βαββ

αβαα βααα

ββαα αββα

ββαβ βββα

• We see that we can have a total of 12 transitions for each spin (the AA’ part or the BB’ part). However, some of the energies are the same (they are degenerate), so the total number of lines comes down to 10 for each, a total of 20!

• In an AA’XX’ we see two sub-spectra, one for the AA’ part and one for the XX’ part. In a AA’BB’ we see everything on the same region of the spectrum.

αααα

symmetric

antisymmetric

N

N

N

HA

HX

HX'

HA'

OMe

HX

HA

Br

HA'

HX'

More than 2 spins (…)More than 2 spins (…)

• Some examples of spin systems giving rise to AA’XX’ and AA’BB’ patterns are given below.

• A typical AA’BB’ spectrum is that of ODCB, orthodichloro benzene . There are so many signals and they are so close to

Cl

HA

HB

HB'

HA'

Cl

benzene . There are so many signals and they are so close to each other, that this compound is used to calibrate instrument resolution.

Common cases for 2Common cases for 2ndnd order systemsorder systems

• So what type of systems we will commonly encounter that will give rise to 2nd order patterns? Most of the times, aromatics will give 2nd order systems because the chemical shift differences of several of the aromatic protons will be very close (0.1 to 0.5 ppm), and JHH in aromatics are relatively large (9 Hz for 3J, 3 Hz for 4J, and 0.5 Hz for 5J).

• For other compounds, the general rule is that if protons in similar environments are ‘fixed’ (that is, they have restricted rotation), we will most likely have 2nd order patterns.

• A typical example of this, generally of an ABX system, are pro-R and pro-S (i.e., diastereotopic) protons of methylenes next to a chiral center:

• In this case, we will have two protons thatare coupled to one another (because theyare different, A and B), they will have verysimilar chemical shifts, and can be coupledto other spins. For example, the oxetaneprotons in taxol:

HA HB

R1 R2

PhRO

HA HB

HX HN PhO

O

O

HB

HA

OH